Riemann surface
matlab

Example 1: Solve the system of equations with augmented matrices using the Gaussian elimination with back-substitution method.
$$\begin{array}{r} x-2 y-z=2 \ 2 x-y+z=4 \ -x+y-2 z=-4 \end{array}$$

Solution:
Step 1: Write the system of equations in an augmented matrix
$$\left[\begin{array}{rrr|r} 1 & -2 & -1 & 2 \ 2 & -1 & 1 & 4 \ -1 & 1 & -2 & -4 \end{array}\right]$$
Step 2: Get a 1 in the first row of the first column
Step 3: Use row 1 to get 0 ‘s in the first column of rows 2 and 3
For the second row we can obtain a zero by multiplying row 1 by -2 and adding it to row 2 .
\begin{aligned} & {\left[\begin{array}{rrr|r} 1 & -2 & -1 & 2 \ 2 & -1 & 1 & 4 \ -1 & 1 & -2 & -4 \end{array}\right]-2 R_1+R_2} \ & -2\left[\begin{array}{llll} 1 & -2 & -1 & 2 \end{array}\right]=\left[\begin{array}{lll|l} -2 & 4 & 2 & -4 \end{array}\right] \ & =\frac{\left[\begin{array}{rrr|r} 2 & -1 & 1 & 4 \end{array}\right]}{=\left[\begin{array}{lll|l} 0 & 3 & 3 & 0 \end{array}\right]} \ & {\left[\begin{array}{rrr|r} 1 & -2 & -1 & 2 \ 0 & 3 & 3 & 0 \ -1 & 1 & -2 & -4 \end{array}\right]} \ & \end{aligned}

Example 1 (Continued):
For the third row we can simply add row 1 to row 3.
\begin{aligned} & \left.\left[\begin{array}{rrr|r} 1 & -2 & -1 & 2 \ 0 & 3 & 3 & 0 \ -1 & 1 & -2 & -4 \end{array}\right] R_1+R_3 \quad \begin{array}{lrr|r} 1 & -2 & -1 & 2 \end{array}\right] \ & {\left[\begin{array}{rrr|r} 1 & -2 & -1 & 2 \ 0 & 3 & 3 & 0 \ 0 & -1 & -3 & -2 \end{array}\right]} \end{aligned}

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